In Calculus- the height of an object in meters after t seconds is given by H(t) = 320t - 16t^2 and H'(t) = 320 - 32t. What is the height after 1 second and what is the velocity after 1 second? when does it hit the ground? and what is the velocity when it hits. please explain.

Question

In Calculus- the height of an object in meters after t seconds is given by H(t) = 320t - 16t^2 and H'(t) = 320 - 32t.   What is the height after 1 second and what is the velocity after 1 second? when does it hit the ground? and what is the velocity when it hits. please explain.

loser66 · Answer

@e.mccormick

e.mccormick · Answer

Height is a position function.  The first derivative of position is velocity, the second is acceleration, and the third is jerk.  For this you just need the position and velocity, which they have given you.

e.mccormick · Answer

As for when it hits the ground, solev for height = 0.  Find the time, then plug the time into the velocity.

loser66 · Answer

@e.mccormick hey friend, that doesn't help at all. I can understand what you mean but pleaaaase, give more explanation

e.mccormick · Answer

From the question:
...height... given by H(t) = 320t - 16t^2
So put in 1 and you have the height after 1 second.
You have the fist derivative: H'(t) = 320 - 32t
Put in 1 and you have the velocity after 1 second.  Those are the first answers.

Then, set H(t) = 320t - 16t^2 equal to 0.  So 0 = 320t - 16t^2 and solve for t.  It must give a positive answer because time does not flow backwards, so discard any negative results.

Once you have the new t, put it into  H'(t) = 320 - 32t and you have the velocity at impact.

anonymous · Answer

thank you! you helped clear it up for me!

loser66 · Answer

:)